Question

ᴋᴏɪ ᴇᴋ ᴀɴsᴡᴇʀ ᴋʀᴅᴏ ʏʀʀ....ᴅᴏɴᴛ sᴘᴀᴍ ᴘʟs ​

Answer 1

Construction: Join PB and extend it to meet CD produced at R.

To prove: PQ∥AB and PQ= 1/2 (AB+DC)

Proof : In △ABP and ΔDRP,

∠APB=∠DPR (Vertically opposite angles)

∠PDR=∠PAB (Alternate interior angles are equal)

AP=PD(P is the mid point of AD)

Thus, by ASA congruency,

ΔABP≅ΔDRP.

By CPCT,PB=PR and AB=RD

lnΔBRC

Q is the mid point of BC (Given)

P is the mid point of BR

(AsPB=PR)

So, by midpoint theorem, PQ ∥RC

⇒PQ∥DC

But AB∥DC(Given)

So,PQ∥AB

Also,PQ=1/2 (RC)…. (using midpoint theorem)

PQ= 1/2 (RD+DC)

PQ= 1/2 (AB+DC)(∵AB=RD)

Hence proved

$\huge\mathfrak\star\pink{hope\:it\:helps}\star$

Answer 2

Answer:

Construction: Join PB and extend it to meet CD produced at R.

To prove: PQ∥AB and PQ= 1/2 (AB+DC)

Proof : In △ABP and ΔDRP,

∠APB=∠DPR (Vertically opposite angles)

∠PDR=∠PAB (Alternate interior angles are equal)

AP=PD(P is the mid point of AD)

Thus, by ASA congruency,

ΔABP≅ΔDRP.

By CPCT,PB=PR and AB=RD

lnΔBRC

Q is the mid point of BC (Given)

P is the mid point of BR

(AsPB=PR)

So, by midpoint theorem, PQ ∥RC

⇒PQ∥DC

But AB∥DC(Given)

So,PQ∥AB

Also,PQ=1/2 (RC)…. (using midpoint theorem)

PQ= 1/2 (RD+DC)

PQ= 1/2 (AB+DC)(∵AB=RD)

Hence proved

\huge\mathfrak\star\pink{hope\:it\:helps}\star⋆hopeithelps